Lesson Objectives

- Demonstrate an understanding of Algebraic Expressions
- Demonstrate an understanding of the Multiplicative Inverse Property
- Demonstrate an understanding of the Multiplicative Identity Property
- Learn about the Multiplication Property of Equality
- Learn how to solve an equation using Multiplication
- Learn how to solve an equation using Division

## How to Solve Equations using Multiplication and Division

In our last lesson, we learned how to solve equations using addition or subtraction. We learned a property known as the addition property of equality, which allows us to add or subtract the same value to or from both sides of an equation. In an equation such as:

x - 5 = 7 Our goal is to isolate x, here 5 is being subtracted away from x. We can add 5 to the left side of the equation to get rid of the - 5 part:

x - 5 + 5 » x + 0 » x In order to make this legal and balance the equation, we must also add the same value of 5 to the other side of the equation:

7 + 5 » 12

When we put the two sides together: x - 5 + 5 = 7 + 5 x = 12

We see our solution here is x = 12, we can check the result by plugging in a 12 for x and evaluating:

x - 5 = 7

12 - 5 = 7

7 = 7

Since the same value (7) is on both sides of the equation, we know our solution (x = 12) is correct. In this lesson, we will now focus on how to solve equations using multiplication and division.

3 • 7 = 21

If we then divide 21 by 7, we are back to 3:

21 ÷ 7 = 3

Let's suppose we encountered an equation such as:

2x = 10

How can we solve this equation? Essentially 2 multiplied by some unknown value is equal to 10. In other words:

2 • ? = 10

We know we can use a related division statement here and determine that our unknown number is 5, however, we want a step-by-step process. When solving an equation, our goal is to isolate x on one side of the equation. On the left side 2 is multiplying x, so in order to undo that operation, we use division: $$2x=10$$ $$\frac{2x}{2}=\frac{10}{2}$$ Dividing both sides of the equation by 2 isolates x on the left side: $$\require{cancel}\frac{\cancel{2}x}{\cancel{2}}=\frac{\cancel{10}5}{\cancel{2}}$$ $$x=5$$ We can check our result by plugging in a 5 for x in our original equation:

2x = 10

2(5) = 10

10 = 10

Since both sides are the same value (10), we know our solution (x = 5) is correct. Let's take a look at a few examples.

Example 1: Solve each equation.

-7x = 21

To solve this equation, we need to isolate x. On the left side, -7 is multiplying x. We can undo this multiplication by dividing both sides of the equation by -7: $$\frac{-7x}{-7}=\frac{21}{-7}$$ $$\frac{\cancel{-7}x}{\cancel{-7}}=\frac{-3\cancel{21}}{\cancel{-7}}$$ $$x=-3$$ We can check our result by plugging in a (-3) for x in our original equation:

-7x = 21

-7(-3) = 21

21 = 21

Since both sides are the same value (21), we know our solution (x = -3) is correct.

Example 2: Solve each equation.

$$\frac{x}{9}=17$$ To solve this equation, we need to isolate x. On the left side, x is being divided by 9. We can undo the division by multiplying both sides of the equation by 9: $$9 \cdot \frac{x}{9}=9 \cdot 17$$ $$\cancel{9}\cdot \frac{x}{\cancel{9}}=9 \cdot 17$$ $$x=153$$ We can check our result by plugging in a 153 for x in our original equation:

$$\frac{x}{9}=17$$ $$\frac{153}{9}=17$$ $$17=17$$ Since both sides are the same value (17), we know our solution (x = 153) is correct.

Example 3: Solve each equation. $$\frac{3}{5}x=-39$$ To solve this equation, we need to isolate x. On the left side, x is being multiplied by the fraction 3/5. We can undo the multiplication with division. We can divide each side by the fraction 3/5, however it is quicker to just multiply both sides by the reciprocal of 3/5, which is 5/3. Remember a number such as 3/5 multiplied by its reciprocal is 1: $$\frac{5}{3}\cdot \frac{3}{5}x=\frac{5}{3}\cdot -39$$ $$\frac{\cancel{5}}{\cancel{3}}\cdot \frac{\cancel{3}}{\cancel{5}}x=\frac{5}{\cancel{3}}\cdot -13\cancel{39}$$ $$x=-13 \cdot 5$$ $$x=-65$$ We can check our result by plugging in a (-65) for x in our original equation: $$\frac{3}{5}x=-39$$ $$\frac{3}{5}\cdot -65=-39$$ $$\frac{3}{\cancel{5}}\cdot -13\cancel{65}=-39$$ $$3 \cdot -13=-39$$ $$-39=-39$$ Since both sides are the same value (-39), we know our solution (x = -65) is correct.

x - 5 = 7 Our goal is to isolate x, here 5 is being subtracted away from x. We can add 5 to the left side of the equation to get rid of the - 5 part:

x - 5 + 5 » x + 0 » x In order to make this legal and balance the equation, we must also add the same value of 5 to the other side of the equation:

7 + 5 » 12

When we put the two sides together: x - 5 + 5 = 7 + 5 x = 12

We see our solution here is x = 12, we can check the result by plugging in a 12 for x and evaluating:

x - 5 = 7

12 - 5 = 7

7 = 7

Since the same value (7) is on both sides of the equation, we know our solution (x = 12) is correct. In this lesson, we will now focus on how to solve equations using multiplication and division.

### Multiplication Property of Equality

Now that we understand the basic principles behind solving an equation with addition and subtraction, it is time to learn how to solve an equation using multiplication and division. Before we discuss the multiplication property of equality, let’s recall two properties from pre-algebra. First and foremost, we have a property known as the multiplicative inverse property. This tells us that a non-zero number multiplied by its inverse (reciprocal) always results in 1: $$\frac{3}{5}\cdot \frac{5}{3}=1$$ $$7 \cdot \frac{1}{7}=1$$ The second property is known as the multiplicative Identity property. This property tells us that multiplying a number by 1 leaves the number unchanged: $$6 \cdot 1=6$$ $$-\frac{2}{9}\cdot 1=-\frac{2}{9}$$ These properties seem fairly obvious in this context, but when solving equations, they may not seem so obvious. Lastly, let's discuss the multiplication property of equality. The Multiplication Property of Equality tells us that we can multiply or divide both sides of an equation by the same non-zero number and not change the solution. Again, just like with the addition property of equality, we have to ensure we perform the same action to both sides of the equation. We need to also make sure we understand that multiplication and division are inverse operations. This means multiplication will undo division and vice versa. Suppose we had:3 • 7 = 21

If we then divide 21 by 7, we are back to 3:

21 ÷ 7 = 3

Let's suppose we encountered an equation such as:

2x = 10

How can we solve this equation? Essentially 2 multiplied by some unknown value is equal to 10. In other words:

2 • ? = 10

We know we can use a related division statement here and determine that our unknown number is 5, however, we want a step-by-step process. When solving an equation, our goal is to isolate x on one side of the equation. On the left side 2 is multiplying x, so in order to undo that operation, we use division: $$2x=10$$ $$\frac{2x}{2}=\frac{10}{2}$$ Dividing both sides of the equation by 2 isolates x on the left side: $$\require{cancel}\frac{\cancel{2}x}{\cancel{2}}=\frac{\cancel{10}5}{\cancel{2}}$$ $$x=5$$ We can check our result by plugging in a 5 for x in our original equation:

2x = 10

2(5) = 10

10 = 10

Since both sides are the same value (10), we know our solution (x = 5) is correct. Let's take a look at a few examples.

Example 1: Solve each equation.

-7x = 21

To solve this equation, we need to isolate x. On the left side, -7 is multiplying x. We can undo this multiplication by dividing both sides of the equation by -7: $$\frac{-7x}{-7}=\frac{21}{-7}$$ $$\frac{\cancel{-7}x}{\cancel{-7}}=\frac{-3\cancel{21}}{\cancel{-7}}$$ $$x=-3$$ We can check our result by plugging in a (-3) for x in our original equation:

-7x = 21

-7(-3) = 21

21 = 21

Since both sides are the same value (21), we know our solution (x = -3) is correct.

Example 2: Solve each equation.

$$\frac{x}{9}=17$$ To solve this equation, we need to isolate x. On the left side, x is being divided by 9. We can undo the division by multiplying both sides of the equation by 9: $$9 \cdot \frac{x}{9}=9 \cdot 17$$ $$\cancel{9}\cdot \frac{x}{\cancel{9}}=9 \cdot 17$$ $$x=153$$ We can check our result by plugging in a 153 for x in our original equation:

$$\frac{x}{9}=17$$ $$\frac{153}{9}=17$$ $$17=17$$ Since both sides are the same value (17), we know our solution (x = 153) is correct.

Example 3: Solve each equation. $$\frac{3}{5}x=-39$$ To solve this equation, we need to isolate x. On the left side, x is being multiplied by the fraction 3/5. We can undo the multiplication with division. We can divide each side by the fraction 3/5, however it is quicker to just multiply both sides by the reciprocal of 3/5, which is 5/3. Remember a number such as 3/5 multiplied by its reciprocal is 1: $$\frac{5}{3}\cdot \frac{3}{5}x=\frac{5}{3}\cdot -39$$ $$\frac{\cancel{5}}{\cancel{3}}\cdot \frac{\cancel{3}}{\cancel{5}}x=\frac{5}{\cancel{3}}\cdot -13\cancel{39}$$ $$x=-13 \cdot 5$$ $$x=-65$$ We can check our result by plugging in a (-65) for x in our original equation: $$\frac{3}{5}x=-39$$ $$\frac{3}{5}\cdot -65=-39$$ $$\frac{3}{\cancel{5}}\cdot -13\cancel{65}=-39$$ $$3 \cdot -13=-39$$ $$-39=-39$$ Since both sides are the same value (-39), we know our solution (x = -65) is correct.

#### Skills Check:

Example #1

Solve each equation. $$-4x=20$$

Please choose the best answer.

A

$$x=3$$

B

$$x=-5$$

C

$$x=-3$$

D

$$x=5$$

E

$$x=-\frac{5}{4}$$

Example #2

Solve each equation. $$\frac{7}{5}x=49$$

Please choose the best answer.

A

$$x=-21$$

B

$$x=-7$$

C

$$x=-14$$

D

$$x=35$$

E

$$x=21$$

Example #3

Solve each equation. $$13x - 9x=-16$$

Please choose the best answer.

A

$$x=-1$$

B

$$x=4$$

C

$$x=12$$

D

$$x=-4$$

E

$$x=-12$$

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